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@ -16,13 +16,13 @@ There are numerous reasons why this project is relevant, as well as a wide range
## 3. The consequences of collisions ## 3. The consequences of collisions
Firstly, it is essential to comprehend how the velocity of the particles after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian). Firstly, it is essential to comprehend how the velocity of the bodies after collision is calculated. We will solve the problem in the elastic collision model (one-dimensional Newtonian).
### 3.1 With particle ### 3.1 With body
Consider two particles, designated as particles `1` and `2`, with respective masses $m_1$ and $m_2$. Before the collision, the velocities of particles 1 and 2 are $v_1$ and $v_2$, respectively. After the collision, the velocities of particles 1 and 2 are $v_1'$ and $v_2'$, respectively. Consider two bodies, designated as bodies `1` and `2`, with respective masses $m_1$ and $m_2$. Before the collision, the velocities of bodies 1 and 2 are $v_1$ and $v_2$, respectively. After the collision, the velocities of bodies 1 and 2 are $v_1'$ and $v_2'$, respectively.
In any collision, momentum is conserved. A collision between to particles is said to be elastic if it involves no change in their internal state. Accordingly, when the law of conservation of energy is applied to such a collision, the internal energy of the particles may be neglected.[^2] In any collision, momentum is conserved. A collision between to bodies is said to be elastic if it involves no change in their internal state. Accordingly, when the law of conservation of energy is applied to such a collision, the internal energy of the bodies may be neglected.[^2]
The conservation of momentum before and after the collision is expressed by the following equation: The conservation of momentum before and after the collision is expressed by the following equation:
@ -87,7 +87,7 @@ m_1v_1 + m_2v_2 = m_1(v_2 + v'_2 - v_1) + m_2v'_2
\end{cases} \end{cases}
$$ $$
From this, we can derive the velocity of the particles following the collision as follows: From this, we can derive the velocity of the bodies following the collision as follows:
$$ $$
\begin{cases} \begin{cases}
@ -98,11 +98,11 @@ $$
### 3.2 With wall ### 3.2 With wall
Consider particle and a wall with respective masses $m$ and $m_w \rightarrow \infty$. Before the collision, the velocities of the particle and the wall are $v$ and $v_w = 0$, respectively. After the collision, the velocities of the particle and the wall are $v'$ and $v_w'$, respectively. Consider bodie and a wall with respective masses $m$ and $m_w \rightarrow \infty$. Before the collision, the velocities of the bodie and the wall are $v$ and $v_w = 0$, respectively. After the collision, the velocities of the bodie and the wall are $v'$ and $v_w'$, respectively.
In order to proceed, we will utilize the formulas derived in section **3.1**. In order to proceed, we will utilize the formulas derived in section **3.1**.
The velocity of the particle will take the following form: The velocity of the bodie will take the following form:
$$ $$
v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v v' = \lim\limits_{m_w \to \infty} \frac{2m_wv_w + v(m-m_w)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{v(m-m_w)}{m+m_w} = -v
@ -114,7 +114,7 @@ $$
v_w' = \lim\limits_{m_w \to \infty} \frac{2mv + v_w(m_w-m)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{2mv}{m+m_w} = 0 v_w' = \lim\limits_{m_w \to \infty} \frac{2mv + v_w(m_w-m)}{m+m_w} = \lim\limits_{m_w \to \infty} \frac{2mv}{m+m_w} = 0
$$ $$
This leads to the conclusion that the wall will not change its position, but the particle will change its velocity value to the opposite. This leads to the conclusion that the wall will not change its position, but the bodie will change its velocity value to the opposite.
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